这可能是一个不起眼的问题,但我找不到办法! 我需要在python中清除stdin缓冲区。
想象一下,我正在运行以下bash脚本:
i=0
for (( ; ; ))
do
echo "$i"
((i++))
done
从命令行运行如下:./loop.sh | python myProg.py
在myProg.py中我希望得到如下内容:
count = 100
f = fileinput.input()
while True:
sleep(2)
# clear stdin somehow ...
# and read the most recent 100 lines
i = 0
while i < count:
myBuffer[i] = f.readline()
if len(myBuffer[i]) > 0:
i += 1
print myBuffer
我不认为我可以在它之前阅读所有的线条,因为它以高速率吐出它们,如果睡眠(仅用于测试atm)是几分钟,它似乎很傻......是有一种方法在python中设置stdin缓冲区大小?或者只是截断/清除它?顺便说一下我使用python 2所以没有bufsize参数
我看了How to avoid Python fileinput buffering
任何python方式吗?但也可以尝试不缓冲:https://unix.stackexchange.com/questions/25372/turn-off-buffering-in-pipe
更新: unbuffer或stdbuf没有运气......
答案 0 :(得分:0)
以防其他人有这个问题,我一直在进行广泛的搜索并得出结论,由于内核实现等原因,这是不可能的......
我写了下面的工作来做我想做的事。我创建了两个文件:textFilter.py和getLatest.py。基本上你运行./loopPrinting.sh | python textFilter.py并获得最新的100行。我有理由相信这是原子和可靠的(虽然如果不是请告诉我!!!)。它创建了16个文件并更改了计数(类似于我认为的苹果直播)。
textFilter.py
import sys
import os
import time
def mainLoop(size):
myBuffer = [0]*size
count = 0
while True:
for i in range(size):
myBuffer[i] = sys.stdin.readline()
f = open('/home/development/textFilter/' + repr(count) + '.dat', 'w')
f.write(''.join(myBuffer))
f.close()
f = open('/home/development/textFilter/count.dat~', 'w')
f.write(repr(count))
f.flush()
os.fsync(f.fileno())
f.close()
time.sleep(0.01)
os.rename('/home/development/textFilter/count.dat~','/home/development/textFilter/count.dat')
count += 1
count = count % 16
if __name__ == "__main__":
try:
mainLoop(int(sys.argv[1]))
except Exception:
mainLoop(100)
getLatest.py
import sys
import time
def getData():
f = open('count.dat', 'r')
count = f.read()
f.close()
count = int(count)
oldCount = count
while oldCount == count:
time.sleep(0.1)
f = open('count.dat', 'r')
count = f.read()
f.close()
count = int(count)
f = open(repr(count) + '.dat', 'r')
data = f.readlines()
f.close()
for row in data:
print row,
return 0
if __name__ == "__main__":
try:
getData()
except Exception as inst:
sys.stderr.write('ERROR')
sys.stderr.write(inst)